Question

I use the Airplay SDK which is a platform for building C++ apps for smartphones. It also has a x86 simulator which uses MS Visual C++ IDE + compiler.

Now, I have this class :

namespace Fair {

    class Bitmap : public Sprite {

    public:

        const CIw2DImage* const& getBitmapData() { return bitmapData; }; // warning: returning reference to temporary

    private:

        CIw2DImage* bitmapData;
    };

}

I get the above warning if I build with GCC (ARM) Debug. I don’t get a warning with (x86) Debug.

I asked elsewhere and I got this reply :

Because `const CIw2DImage* const’ is a
const pointer to const CIw2DImage, and
Bitmap::bitmapData is a pointer to
non-const CIw2DImage compiler
automatically casts pointer to
non-const to const, so here’s a
temporary. The following code might be
generated by a “typical” compiler:

const CIw2DImage* const& getBitmapData() {
    const CIw2DImage* const tmp = bitmapData;
    return tmp;
}

Probably (x86) compiler doesn’t detect
this problem.

You might want to remove reference
symbol (&) from the prototype (why do
you want use a reference in this
case?)

If a compiler does that, then it’s totally wrong practice..? Making the value returned more “strict” is simply at compiler-level, to prevent “abuse”. (x86) doesn’t detect because it doesn’t “cause” the problem in the 1st case..?

I return a reference to a pointer for the sole reason to “save” 32 bits of memory, i.e. use the same block of memory as the bitmapData pointer but within a different context.

Any comments please?

Answer 1

You won’t save any memory at all returning a reference, it’s just a pointer internally. In addition, const occurs completely at compile-time, there’s no reason for any temporary to be produced- and even if it was, a const reference to it would be valid.

That code looks terrible to me personally but it’s not undefined in any way.